1. Chapter 15 Class 11 Statistics
2. Serial order wise

Transcript

Misc 2 The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations. Let the other two observations be x and y. Therefore, our observations are 2, 4, 10, 12, 14, x, y. Given Mean = 8 i.e. ๐๐ข๐ ๐๐ ๐๐๐ ๐๐๐ฃ๐๐ก๐๐๐๐ ๏ทฎ๐๐ข๐๐๐๐ ๐๐ ๐๐๐ ๐๐๐ฃ๐๐ก๐๐๐๐ ๏ทฏ = 8 2 + 4 + 10 + 12 + 14 + ๐ฅ + ๐ฆ๏ทฎ7๏ทฏ = 8 42 + x + y = 7 ร 8 x + y = 56 โ 42 x + y = 14 Also, Given Variance = 16 1๏ทฎn๏ทฏ ๏ทฎ๏ทฎ( ๐ฅ๏ทฎ๐๏ทฏ๏ทฏโ ๐ฅ๏ทฏ)๏ทฎ2๏ทฏ = 16 1๏ทฎ7๏ทฏ ๏ทฎ๏ทฎ( ๐ฅ๏ทฎ๐๏ทฏ๏ทฏโ8)๏ทฎ2๏ทฏ = 16 1๏ทฎ7๏ทฏ [(2 โ 8)2 +(4 โ 8)2+(10 โ 8)2+(12 โ 8)2 +(14 โ 8)2+(x โ 8)2 +(y โ 8)2] = 16 1๏ทฎ7๏ทฏ [ (โ6)2 + (โ4)2 + (2)2 + (4)2 + (6)2 + (x โ 8)2 + (y โ 8)2 ] = 16 1๏ทฎ7๏ทฏ [36 + 16 + 4 + 16 + 36 + x2 + (8)2 - 2(8)x + y2 + (8)2 - 2(8)y] = 16 [ 108 + x2 + 64 โ 16x + y2 + 64 โ 16y] = 16 ร 7 [ 236 + x2 + y2 โ 16y โ 16x ] = 112 [ 236 + x2 + y2 โ 16(x + y) ] = 112 [ 236 + x2 + y2 โ 16(14) ] = 112 236 + x2 + y2 โ 224 = 112 x2 + y2 = 112 โ 236 + 224 x2 + y2 = 100 From (1) x + y = 14 Squaring both sides (x + y)2 = 142 x2 + y2 + 2xy = 196 100 + 2xy = 196 2xy = 196 โ 100 2xy = 96 xy = 1๏ทฎ2๏ทฏ ร 96 xy = 48 x = 48๏ทฎ๐ฆ๏ทฏ Putting (3) in (1) x + y = 14 48๏ทฎ๐ฆ๏ทฏ + y = 14 48 + y2 = 14y y2 โ 14y + 48 = 0 y2 โ 6y โ 8y + 48 = 0 y(y โ 6) โ 8(y โ 6) = 0 (y โ 6)(y โ 8) = 0 So, y = 6 & y = 8 For y = 6 x = 48๏ทฎ๐ฆ๏ทฏ = 48๏ทฎ6๏ทฏ = 8 Hence x = 8, y = 6 are the remaining two observations For y = 8 x = 48๏ทฎ๐ฆ๏ทฏ = 48๏ทฎ8๏ทฏ = 6 Hence, x = 6, y = 8 are the remaining two observations Thus, remaining observations are 6 & 8

Serial order wise